Given the matrix game $G$, we construct the $k$-player game ${\mathcal G}$
in which the $i$th player's strategy space is the set of all
probability distribution functions over $S_i$ and the payoff is given
by the personalized payoff function defined above.  
We can view the strategy space as the set of probability distribution functions over $S_i$ instead of weight assignments to $E_i$ since a weight assignment uniquely defines a probability distribution function, and since the payoffs and responses of the other players only depend on the $p_i(s)$ values, not on the $w_i(e)$ values.
Then a
personalized equilibrium of $G$ is equivalent to a Nash equilibrium of
${\mathcal G}$.  By~\cite[Proposition~20.3]{OsborneRubinstein94}, a game has
a pure Nash equilibrium if the strategy space of each player is a
compact, non-empty, convex space, and the payoff function of each
player is continuous on the strategy space of all players and
quasi-concave in the strategy space of that player.  The set of
probability distributions over $S_i$ is clearly nonempty, convex, and
compact.  Furthermore, given probability distributions $p_i$ over
$S_i$, $1 \le i \le k$, the payoff for any player $i$ is simply a
solution to the following linear program with variables $w_i(e)$, over
$e \in E_i$.
\begin{eqnarray*}
\max \sum_{e\in E_i} w_i(e)u_i(e) && \\
\sum_{e \in E_i: s \in e} w_i(e) & \leq p_j(s) & s \in S_j, 1\le j \le k\\
\sum_{e \in E_i} w_i(e) & = 1 & \\ 
w_i(e) & \ge 0 & e \in E
\end{eqnarray*}
It is easy to see that the payoff function is both continuous in the
probability distributions of all players, and quasi-concave in the
strategy space of player $i$, thus completing the proof of the
theorem.